A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-30 m/s)i and a second piece, also of mass m, moves with velocity (-30 m/s)j. The third piece has mass 3m.

(a) Just after the explosion, what is the magnitude of the velocity of the third piece?
(b) What is its direction at this time?
° (counterclockwise from the +x axis)

The momentums have to add to zero.

ah ok thanks?

from what i understand when I solved this problem is that the masses cancel out each other anyway by the end, so just ignore them. Because the two velocities that they give you are vertical and horizontal components square them both and add them and take the square root. Divide by 3 because the third piece is 3m.

(-31)i+ (-31)j+3v=0

To get the direction draw a triangle with the vertical line being 30 and the horizontal line being 30 since they are the horizontal and vertical components of the velocity. Then use tan-1 (30/30) to find the angle.

woops, i put -31 for the equation, i meant to put -30.
oh and the momentums have to add to zero because of the conservation of linear momentum. The initial momentum is 0 since the velocity is 0.

To solve the problem correctly, let's start by considering the conservation of linear momentum. According to this principle, the total momentum before the explosion must be equal to the total momentum after the explosion.

The initial momentum is zero since the vessel is at rest. Therefore, the total momentum after the explosion should also be zero.

Let's consider the components of momentum in the x and y directions separately.

In the x-direction, the first piece has a velocity of (-30 m/s)i, and the second piece has a velocity of 0 m/si (since it only has a component in the y-direction). The third piece has an unknown velocity, which we can call (vx m/s)i.

So, the x-component of momentum after the explosion is given by the sum of the x-components of the individual pieces:

(-30 m/s)m + 0 m/s + (vx m/s)(3m) = 0

Simplifying the equation:

-30m + 0 + 3vx m^2/s^2 = 0

Now, let's consider the y-component of momentum. The first piece does not have any y-component of velocity, so it does not contribute in the y-direction. The second piece has a velocity of (-30 m/s)j, and the third piece has an unknown velocity, which we can call (vy m/s)j.

So, the y-component of momentum after the explosion is given by the sum of the y-components of the individual pieces:

0 + (-30 m/s)m + (vy m/s)(3m) = 0

Simplifying the equation:

-30m + 3vy m^2/s^2 = 0

Now, we have a system of two equations:

-30m + 0 + 3vx m^2/s^2 = 0
-30m + 3vy m^2/s^2 = 0

To solve this system, we can start by eliminating one of the unknowns. Let's eliminate vy by multiplying the first equation by 3 and the second equation by -1:

-90m + 9vx m^2/s^2 = 0
30m - 3vy m^2/s^2 = 0

By adding these two equations together, we get:

-90m + 9vx m^2/s^2 + 30m - 3vy m^2/s^2 = 0

Simplifying the equation:

-60m = -6vx m^2/s^2 + 6vy m^2/s^2

Dividing both sides of the equation by -6m, we obtain:

10 = -vx + vy

Now, we have an equation relating the x and y components of the velocity of the third piece.

To find the magnitude of the velocity, we can use the Pythagorean theorem:

|v|^2 = vx^2 + vy^2

Substituting the equation we obtained earlier:

|v|^2 = (-vx + 10)^2 + vx^2

Expanding and simplifying the equation:

|v|^2 = vx^2 - 20vx + 100 + vx^2

Combining like terms:

|v|^2 = 2vx^2 - 20vx + 100

We want to find the magnitude of the velocity, so |v| = √(|v|^2). Substituting the equation above:

|v| = √(2vx^2 - 20vx + 100)

Finally, to find the direction of the velocity, we can use trigonometry. The angle counterclockwise from the +x axis can be found using the arctan function:

θ = tan^(-1)(vy/vx)

Substituting the equation relating vy and vx obtained earlier:

θ = tan^(-1)((-vx + 10)/vx)

Now, you can use these equations to solve for the magnitude and direction of the velocity of the third piece.